Triangulation of 4 points why Delaunay maximizes the minimum? I have been going through the chapter on Delaunay triangulations from the book by DeBerg (http://www.cs.uu.nl/geobook/interpolation.pdf). In lemma 9.4, he simply says that "from Thales theorem" we can show that of the two possible triangulations of four points (depending on which diagonal you choose to join) the one where we cut through the angles that sum to less than 180 is "illegal" meaning the angles don't satisfy the maximin criterion (meaning of the six resulting angles of the triangulation, the minimum is not as high as it would have been if we had joined the other diagonal). I don't really see how one can get this from Thales theorem. Does any one have an outline for a proof as to why this might hold? 
 A: I guess that by “Thales' theorem” he's essentially referring to the inscribed angle theorem, probably in the form of Theorem 9.2 from that chapter.
Consider circles through $p_i$ and $p_k$. Concentrate on the arc above that line. Any point on such a circle will form the same angle with $p_i$ and $p_k$. The centers of all these circles will lie on the perpendicular bisector between $p_i$ and $p_k$. As the center of a circle moves up, the angle becomes smaller, while increasing angle corresponds to lower center. My point is this: in order to maximize an angle, you want to move the center of the corresponding circle as far down as you can.

So from all the circles I just described, there are two of special importance, namely the one through $p_j$ and through $p_l$. Of these two, the one through $p_l$ is the one which has the lower center position, so it corresponds to a larger angle.
$$\angle p_kp_lp_i > \angle p_kp_jp_i$$
So $\triangle p_kp_lp_i$ is the combination you want if you maximize the angle opposite side $p_ip_k$.
So far, this is comparing just two angles. Couldn't it be the case that switching from edge $p_ip_j$ to edge $p_kp_l$ decreases some other angle instead? The answer is no:
\begin{align*}
\angle p_kp_ip_l = \angle p_kp_ip_j + \angle p_jp_ip_l &> \angle p_kp_ip_j \\
\angle p_lp_jp_k = \angle p_lp_jp_i + \angle p_ip_jp_k &> \angle p_ip_jp_k
\end{align*}
So for both the other angles you're adding something, therefore they will increase not decrease. This is due to the quadrilateral being convex. The only angle which actually decreases is the angle at $p_k$, since that will get divided into two. To see that this is still an improvement, you'd do the whole consideration above, but this time for edge $p_lp_i$ instead of $p_ip_k$.
